In a formation equation, the standard state of elements combined together to form 1 mole of product. Here in this example it says write the formation equation for methane thiol which is chapter 3, SH. Following the steps, it says step one, write the compound as a product. So chapter 3 SH. In a formation equation, the product will always have a coefficient of one because remember, we're making one mole of it.
Step 2. Write the standard states of the elements that make up the element as the reactants. So it's made-up of carbon. Carbon is monoatomic, and the most common form of carbon is graphite plus hydrogen. Hydrogen is diatomic, and it's a gas plus sulfur. Sulfur is polyatomic. It's eight and it is a solid. Write the formation equation with the appropriate coefficients for the reactants.
A formation equation is a rare instance where a coefficient doesn't need a whole number. So if we were to balance this out, remember we make a list on both sides. So this is carbon, hydrogen and sulfur. Carbon, hydrogen and sulfur. We have one carbon, 2 hydrogens, 8 sulfurs, 1 carbon, 4 hydrogens, 1 sulfur. The coefficient here has to be a one. We cannot change that, so we're going to have to adjust the other coefficients.
Your carbons are tied for both 1:00 and 1:00, so that's fine. Here we have 4 hydrogens, but here we have two. So I'm going to put a 2 here. That means the two gets distributed to give us 4 hydrogens. Now then, we have 8 hydrogens here, but only one here. I'm not allowed to put a coefficient in front of the product, so this is one of those rare occasions where I'm going to have to do 18 because 18 * 8 will give me one. So now both sides of the equation have one solver involved and then we can just say there's a coefficient of one if you want in front of the carbon graphite.
So here this would be our balanced equation, 1 mole of carbon graphite +2 moles of hydrogen gas react with 18 sulfur 8 to produce one mole of methane thiol. So this would be our formation equation for methane thiol.